A bullet of mass 20 grams strikes a target with the velocity hundred metre per second and comes to rest. 50% kinetic energy of the bullet raises its temperature by 25 degree centigrade. calculate increase in the internal energy of the bullet due to rise in temperature and the specific heat of the material of the bullet.
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Increase in the value of internal energy is 499.9 J
Explanation:
According to the statement of the question:
50% of K.E = msθ
1/2 [ 1/2 mV^2] = msθ
Now s = V^2/4θ
s = (100)^2/ 4 x 298
s = 10,000 / 1192 = 8. 389 JKg^-1 K^-1
From first law of thermodynamics:
ΔU = Δq + w
w = pΔv [ Δv = 0]
Thus
ΔU = Δq = msθ
= 20/100 x 8. 389 x 298
ΔU = 499.9 J
Increase in the value of internal energy is 499.9 J
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